1. **State the problem:** We need to evaluate the definite integral $$\int_0^3 \frac{dx}{x + \sqrt{9 - x^2}}.$$\n\n2. **Understand the integral:** The denominator is $x + \sqrt{9 - x^2}$. Note that $\sqrt{9 - x^2}$ represents the upper half of a circle of radius 3 centered at the origin.\n\n3. **Rationalize the denominator:** Multiply numerator and denominator by the conjugate $\sqrt{9 - x^2} - x$ to simplify:\n$$\int_0^3 \frac{dx}{x + \sqrt{9 - x^2}} = \int_0^3 \frac{\sqrt{9 - x^2} - x}{(x + \sqrt{9 - x^2})(\sqrt{9 - x^2} - x)} dx.$$\n\n4. **Simplify the denominator:** Using difference of squares:\n$$ (x + \sqrt{9 - x^2})(\sqrt{9 - x^2} - x) = (\sqrt{9 - x^2})^2 - x^2 = 9 - x^2 - x^2 = 9 - 2x^2.$$\n\n5. **Rewrite the integral:**\n$$\int_0^3 \frac{\sqrt{9 - x^2} - x}{9 - 2x^2} dx = \int_0^3 \frac{\sqrt{9 - x^2}}{9 - 2x^2} dx - \int_0^3 \frac{x}{9 - 2x^2} dx.$$\n\n6. **Evaluate the second integral:** Let’s compute $$I_2 = \int_0^3 \frac{x}{9 - 2x^2} dx.$$\nUse substitution $u = 9 - 2x^2$, then $du = -4x dx$, so $x dx = -\frac{du}{4}$.\nWhen $x=0$, $u=9$; when $x=3$, $u=9 - 2(9) = 9 - 18 = -9$.\n\nSo,\n$$I_2 = \int_{u=9}^{-9} \frac{-1/4}{u} du = -\frac{1}{4} \int_9^{-9} \frac{1}{u} du = -\frac{1}{4} [\ln|u|]_9^{-9} = -\frac{1}{4} (\ln 9 - \ln 9) = 0.$$\n\n7. **Evaluate the first integral:** $$I_1 = \int_0^3 \frac{\sqrt{9 - x^2}}{9 - 2x^2} dx.$$\nUse substitution $x = 3 \sin \theta$, so $dx = 3 \cos \theta d\theta$, and $\sqrt{9 - x^2} = 3 \cos \theta$.\nWhen $x=0$, $\theta=0$; when $x=3$, $\theta=\frac{\pi}{2}$.\n\nRewrite denominator:\n$$9 - 2x^2 = 9 - 2(9 \sin^2 \theta) = 9 - 18 \sin^2 \theta = 9(1 - 2 \sin^2 \theta).$$\n\nIntegral becomes:\n$$I_1 = \int_0^{\pi/2} \frac{3 \cos \theta}{9(1 - 2 \sin^2 \theta)} \cdot 3 \cos \theta d\theta = \int_0^{\pi/2} \frac{9 \cos^2 \theta}{9(1 - 2 \sin^2 \theta)} d\theta = \int_0^{\pi/2} \frac{\cos^2 \theta}{1 - 2 \sin^2 \theta} d\theta.$$\n\n8. **Simplify denominator:** Note $1 - 2 \sin^2 \theta = \cos 2\theta$ (since $\cos 2\theta = 1 - 2 \sin^2 \theta$).\n\nSo,\n$$I_1 = \int_0^{\pi/2} \frac{\cos^2 \theta}{\cos 2\theta} d\theta.$$\n\n9. **Express $\cos^2 \theta$ in terms of double angle:**\n$$\cos^2 \theta = \frac{1 + \cos 2\theta}{2}.$$\n\nTherefore,\n$$I_1 = \int_0^{\pi/2} \frac{\frac{1 + \cos 2\theta}{2}}{\cos 2\theta} d\theta = \frac{1}{2} \int_0^{\pi/2} \left( \frac{1}{\cos 2\theta} + 1 \right) d\theta = \frac{1}{2} \int_0^{\pi/2} \sec 2\theta + 1 d\theta.$$\n\n10. **Split the integral:**\n$$I_1 = \frac{1}{2} \left( \int_0^{\pi/2} \sec 2\theta d\theta + \int_0^{\pi/2} 1 d\theta \right) = \frac{1}{2} \left( \int_0^{\pi/2} \sec 2\theta d\theta + \frac{\pi}{2} \right).$$\n\n11. **Evaluate $\int \sec 2\theta d\theta$:**\nUse substitution $u = 2\theta$, $du = 2 d\theta$, so $d\theta = \frac{du}{2}$.\n\nLimits: when $\theta=0$, $u=0$; when $\theta=\frac{\pi}{2}$, $u=\pi$.\n\nIntegral becomes:\n$$\int_0^{\pi/2} \sec 2\theta d\theta = \int_0^{\pi} \sec u \cdot \frac{du}{2} = \frac{1}{2} \int_0^{\pi} \sec u du.$$\n\n12. **Evaluate $\int_0^{\pi} \sec u du$:**\nThe integral of $\sec u$ is $\ln |\sec u + \tan u| + C$. However, $\sec u$ has a vertical asymptote at $u=\frac{\pi}{2}$, so the integral diverges.\n\n13. **Conclusion:** The integral $$\int_0^3 \frac{dx}{x + \sqrt{9 - x^2}}$$ diverges because the integrand has a vertical asymptote at $x=\sqrt{\frac{9}{2}}$ where the denominator $9 - 2x^2 = 0$, causing the integral to be improper and divergent.